which, by substituting the values of a', b', c', and A', becomes

cos.(180° — A) = cos.(180°—B) cos.(180°C) + sin.(180° -B) sin.(180° C) cos.(180°-a),

But,

(2)

cos.(180°-4)=-cos.A, etc., sin.(180°-B)=sin.B, etc.; and placing these values for their equals in eq. (2), and changing the signs of both members of the resulting equation, we get

cos. A sin.B sin. C cos.a which agrees with the enunciation.

cos.B cos.C,

By treating the other two of formulæ (S), Prop. 7, in the same manner, we should obtain similar values for the cosines of the other two angles of the triangle ABC; or we may get them more easily by a simple permutaLion of the letters A, B, C, a, etc.

Hence, we have the three equations

cos.A sin.B sin. C cos.a

=

cos. B cos. C

cos. A cos.C

(V)

sin. A sin.B cos.c

cos. A cos. B.

From these we can find formula to express the sine or the cosine of one half of the side of a spherical triangle, In terms of the functions of its angles; thus:

Add 1 to each member of eq. (3), and we have

1 + cos.a =

cos.A+cos. B cos. C + sin.B sin. C
sin.B siu. C

and since cos.A + cos.(B-C) · C) = 2cos.}(A + B — C')cos.§ (A+C—B) (Eq. 17, Sec. I, Plane Trig.), we have

2cos.2 a

=

2cos.1(A + B — C')cos.¿(A + C — B)

sin.B sin. C

Make A+B+C=28; then A+ B-C=28—20, A + C— B = 2S—2B, 1(A + B — C) = S— C, and (A +C-B) SB; whence

=

To find the sin.a in terms of the functions of the angles, we must subtract each member of eq. (3) from 1, by which we get

But, 1-cos.a 2sin.'a; hence we have,

2sin.2a=

(sin.B sin. C-cos. B cos. C)-cos. A
sin.B sin.C

Operating upon this in a manner analogous to that by which cos.a was found, we get,

I the first equation in (W) be divided by the first in

(1), we shall have,

(-cos. cos.(S-A) tan.ja={cos.(SB) cos.(S-C

And corresponding expressions may be obtained for tan.b and tan.c.

NAPIER'S ANALOGIES.

If the value of cos.c, expressed in the third equation of group (S), Prop. 7, be substituted for cos.c, in the second member of the first equation of the same group, we have,

cos.acos.a cos.2b+sin.a sin.b cos.b cos. C+ sin.b sin.c cos.A; which, by writing for cos.b its equal, 1—sin.b, becomes, cos.acos.a-cos.a sin.2b+sin.a sin.b cos.b cos. C+sin.b sin.c cos. A. Or, 0 cos.a sin.2b+sin.a sin.b cos.b cos. C+sin.b sin.c cos.A. Dividing through by sin.b, and transposing, we find,

cos. A sin.c cos.a sin.b-sin.a cos.b cos. C;

=

By substituting the value of cos, in the second of the equations of group (S), Prop. 7; or, merely writing B for A, and interchanging b and a, in the above value, for os. A, we obtain,

cos.B=

cos.b sin.a-sin.b cos.a cos. C.
sin.c

(2)

Adding equations (1) and (2), member to member,

we have,

cos.A+cos.B:

sin.(a+b)—sin.(a+b) cos. C ̧

=

sin.c

by remembering that sin.a cos.b+cos.a sin.b≈ sin.(a+b). (See Eq. (7), Sec. I, Plane Trig.).

Whence, cos. A + cos.B= (1—cos. C')

sin.(a+b). (3)

sin.c

In any spherical triangle we have, (Prop. I),

sin.A sin.B: sin.a: sin.b;

And therefore, sin.A+ sin.B: sin.B :: sin.a + sin. :

sin.b.

Hence, sin.A + sin.B_(sin.a + sin.b) sin.B

sin.b

Dividing equation (4) by equation (3), member by

member, we obtain,

=

sin. A+ sin.B sin. C sin.a+sin.b
cos.A+cos.B 1-cos. C sin.(a+b)

(5)

Comparing this equation with Equations (20) and (26), Sec. I, Plane Trigonometry, we see that it can be reduced to

sin.A-sin.B: sin.B :: sin.a-sin.b: sin.b;

Dividing this equation by equation (3), member by member, we obtain,

sin.A-sin.B sin. C sin.a-sin.b

=

cos.A+cos.B 1-cos.C sin.(a+b)

Comparing this with Equations (22) and (26), Sec. I, Plane Trigonometry, we see that it will reduce to

sin.a-sin.b
sin.(a + b)

(7)

́a + b·

a·

Now, sin.a+sin.b2sin.

Sec. I, Plane Trig.).

COS.

; Eq. (15),

2

and, sin. (a + b) = 2sin.(a + 3)

COS.

s.( ĐỎ); Eq (30),

Sec. I, Plane Trig.).

2

a +

2

Dividing the first of these by the second, we have

Writing the second member of this equation for its first member in Eq (6), that equation becomes

tan. (A + B) = cot.

C

cos. (ab)
cos. (a+b)

(8)

By a similar operation, Eq. (7) may be reduced to

tan. }(A — B) = cot. sin. (a—b) (9)

sin. (a+b)

Equations (8) and (9) may be resolved into the proportions

cos. (a + b): cos. 1(a - b) :: cot. 1C: tan. 1(A + B); sin. (a+b) sin. (ab) :: cot. C: tan. (A — B).

:

These proportions are known as Napier's 1st and 2d